The key is the following integral representation:

\begin{equation}
\begin{aligned}
s_n(z)&:= \sum_{j>n} \frac{z^j e^{-j/n}}j \\
&=\sum_{j>n} z^j e^{-j/n} \int_0^\infty du\,e^{-ju} \\
&=\int_0^\infty du\,\sum_{j>n} z^j e^{-j/n}e^{-ju} \\
&=\frac{z^{n+1}}{e^{1+1/n}}\,\int_0^\infty du\,\frac{e^{-(n+1)u}}{1-z e^{-u-1/n}} \\
&=\frac{z^{n+1}}{e^{1+1/n}}\,\int_0^\infty dv\,\frac{e^{-v-v/n}}{n(1-z e^{-(v+1)/n})}.
\end{aligned} \tag{1}
\end{equation}
Next, with $n\ge1$, $|z|=1$, and $v>0$,
\begin{equation*}
|1-z e^{-(v+1)/n}|\ge1-e^{-(v+1)/n}\ge1-\frac1{1+(v+1)/n}=\frac{v+1}{n+v+1},
\end{equation*}
whence
\begin{equation*}
\Big|\frac{e^{-v-v/n}}{n(1-z e^{-(v+1)/n})}\Big|
\le2e^{-v}. \tag{2}
\end{equation*}
Also, for all real $t$ and all real $v>0$,
\begin{equation*}
\Big|\frac{e^{-v}}{1+v-it}\Big|\le e^{-v}. \tag{3}
\end{equation*}

Suppose now that $n\to\infty$ and $z=e^{it/n}$ for some real number $t$ possibly varying with $n$ but staying bounded. Then
\begin{equation*}
\frac{e^{-v-v/n}}{n(1-z e^{-(v+1)/n})}=\frac{e^{-v}}{1+v-it}+o(1)
\end{equation*}
for each real $v>0$.
So, by (1), (2), (3) and dominated convergence,
\begin{align*}
s_n(z)&=
\frac{e^{it}}e\,\int_0^\infty dv\,\frac{e^{-v}}{1+v-it}+o(1)=\Gamma(0,1-i t)+o(1).
\end{align*}

On the other hand, if for $z$ with $|z|=1$ it is not true that $z=e^{it/n}$ for some real number $t$ possibly varying with $n$ but staying bounded, then without loss of generality $n(1-z)\to\infty$, whence
\begin{equation*}
n|1-z e^{-(v+1)/n}|\ge n|1-z|-(v+1)\to\infty
\end{equation*}
for each real $v>0$.
So, by (1), (2), and dominated convergence, $|s_n(z)|\to0$.

So,
\begin{equation*}
-f_n=\min_{|z|=1} \Re s_n(z)\to\mu,
\end{equation*}
where
\begin{equation*}
\mu:=\min_{t>0}r(t),\quad r(t):=\Re\Gamma(0,1-i t).
\end{equation*}
(According to Mathematica's numerics, $\mu=r(t_*)=-0.12686\dots$, where $t_*=2.0287\dots$.)